Talk:Finite promise games
Why should we feel that this function is even well-defined and outgrows all our notations? I'm starting to think that Loader's function, Laver tables and stuff like that is actually hoax. I haven't seen any proofs for their strength, I saw only claims! Ikosarakt1 (talk ^ ) 09:20, June 23, 2014 (UTC) The proof that, for example, Loader's function grows really fast is the fact that the Calculus of Constructions can implement any function provably recursive in higher order logic. So it's like the Busy Beaver function at a lower level. Simple proof that Loader's function outgrows any function provably recursive in higher order logic: let f be such a function, and let n be the number of characters required to implement it. Let m be the number of characters required to implement the doubling function. Then we can output f(2x) using at most m+n+x characters, so for large x we have 2x > m+n+x and so D(2x) >= D(m+n+x) >= f(2x). I don't know the proof that the Calculus of Constructions can implement any function in higher order logic. Anyone have any sources for this? Even proving it for second order arithmetic would be quite interesting. As for Friedman's finite games, I believe Friedman asserted either that the associated function is not provably recursive in ZFC+Mahlo, or that that the function outgrows all functions provably recursive in ZFC+Mahlo. I forget exactly what he said. The latter statement is of course what we are after, but I believe that if the former statement is true we can assume the latter. It is possible for a recursive function to be not provably recursive in a system T but not outgrow all functions provably recursive in T; for example, we could take f to be sufficiently fast growing on even inputs, and zero on odd inputs. But I believe such exceptions to be pathological, so a "naturally defined function" should satisfy the latter statement if it satisfies the former. Of course Friedman may have stated the latter after all, we should look it up. As for Laver tables, I am unsure. We know that there are theorems that are not provable in ZFC+"rank into rank cardinal", but does that mean that the specified function outgrows all functions provably recursive in ZFC+"rank into rank cardinal"? I don't see why this should necessarily be the case, so I am suspicious in the case of Laver tables.Deedlit11 (talk) 13:19, June 23, 2014 (UTC) I just looked up the link, and Friedman does indeed state that the described function dominates all provably recursive functions of SMAH+, which is stronger than ZFC+Mahlo, so there you go. Deedlit11 (talk) 13:22, June 23, 2014 (UTC) :Sorry, but the problem is that our community can't figure out proofs about how these things compares to explicit notations (like up-arrows and \psi ), and these "logics" and "arithmetics" doesn't really give sense of them. How to prove that D(n) indeed outgrows, say, n \uparrow^n n ? Ikosarakt1 (talk ^ ) 13:58, June 23, 2014 (UTC) :: f(n) = n \uparrow^n n is expressible even in Peano arithmetic, so we can write a function in the Calculus of constructions that implements it. So just use the proof above to show that D(n) outgrows f(n). ::Do you accept that the Busy Beaver function outgrows all our recursive functions, even though we can compute only a few values of the Busy Beaver? Then it's the same with these other functions, accept the class of functions that they beat are smaller. If we take the system T, for example, the expressible functions are exactly the functions provably recursive in Peano arithmetic, so the diagonalizing function obviously outgrows f(n). The functions in the fast-growing hierarchy that are provably recursive in Peano arithmetic are exactly the functions below \varepsilon_0 , so I would say we have a good sense of the growth rate of the diagonalizing function of system T. For, say, system F, whose expressible functions are the provably recursive functions of second order arithmetic, the associated ordinal is very, very large, larger than what we can currently express. So I guess for that reason we cannot get a good "sense" of the growth rate. But, it is very likely that all our functions are provably recursive in second order arithmetic, so D(n) will beat all our notations for that reason, even if we cannot get a really good grasp of D(n). Deedlit11 (talk) 14:19, June 23, 2014 (UTC) :::Can you show some more bounds for D(n)? Or can you explain how to do that? Also, can you explain how Friedman's finite games even returns numbers. I don't see any functions here. Wythagoras (talk) 14:54, June 23, 2014 (UTC) ::::I'm actually not that well versed in the Calculus of Constructions. Basically, to get bounds on D(n) is very similar to getting bounds on the Busy Beaver function; to get lower bounds, you present a term with n symbols that generates a large number. Getting upper bounds is harder, you need to evaluate all terms and find the biggest number for a given length. On the plus side, you don't have to worry about nonterminating terms, since every term terminates in the Calculus of Constructions. In David Moews' contest recap, he references an article that describes how to program F_epsilon_0 in the Calculus of Constructions. Obviously this is not that big, but it's a start. ::::For Friedman's finite games, you can read the original post. He gives the theorem ::::PROPOSITION 1.2. Let T be in PL and n >= 1. If m,s are sufficiently large then Bob wins G(T,n,s) even if Bob accepts all factorials > m offered by Alice. ::::So for example we can define f(T,n,m) to be the minimum number S such that if s >= S, Bob wins G(T,n,s) even if Bob accepts all factorials greater than m offered by Alice. There's a bit of difficulty here though, because T is a function from N^k to N, which is equivalent to a real, not a natural number. Perhaps it is the fact that T is piecewise linear, that means that T is reducible to integers. I'm not quite sure. So instead let's use proposition 4.2: ::::PROPOSITION 4.2. Let v_1,...,v_p be in N^k and n,s >= 1. If m is sufficiently large then Bob wins G(v_1,...,v_p;n,s) even if Bob accepts all factorials > m offered by Alice. ::::Define f(v_1,...,v_p;n,s) to be the minimum number M such that if m >= M, Bob wins G(v_1,...,v_p;n,s) even if Bob accepts all factorials > m offered by Alice. Then f is not provably recursive in ZFC+{There exists a k-strong Mahlo}_k. To get a nice unary function, define f(x) to be max f(v_1,...,v_p;n,s) with n, s, and v_ij for all i,j less than or equal to x. Then f(x) dominates all provably recursive functions of ZFC+{There exists a k-strong Mahlo}_k. Deedlit11 (talk) 15:30, June 23, 2014 (UTC) Numbers? I cannot see any statement about functions (or numbers) in the page. How can this theory create large numbers or fast-growing functions? hyp$hyp?cos&cos (talk) 05:01, July 28, 2014 (UTC) :In my opinion, it's just a hoax. Factorials can't provide even tetrational growth rate, but Friedman claims that it grows in the magnitude of something far above f_{\psi(M)}(n) . Ikosarakt1 (talk ^ ) 17:20, July 28, 2014 (UTC) The result is correct, just the article is poorly written and doesn't emphasise what numbers are the ones which are so large. I will try to make article more accessible whenever I'm back home. LittlePeng9 (talk) 17:40, July 28, 2014 (UTC) LittlePeng9 has made a good edit, I would just like to add that if you go to the source page, for the other propositions of the form "For m sufficiently large, S(a,b,c,m)" where S is some statement involving m and some other variables, we can convert that to a fast-growing function by defining f(n) to be the minimum number M so that if m >= M and a, b, and c are of size no greater than n, then S(a,b,c,m) is true. If a,b,c are natural numbers the notion of "size" is obvious, otherwise we have to make a definition (like for polynomials, the "size" is the maximum of the degree and all the coefficients). Deedlit11 (talk) 05:04, July 29, 2014 (UTC) SMAH vs. SMAH+ What's the difference between them? I don't see it. The two statement seem the same. hyp$hyp?cos&cos (talk) 00:18, July 29, 2014 (UTC) :SMAH is ZFC plus, for each \(k\), an axiom of the form "there is a \(k\)-Mahlo cardinal". Note that any proof can only invoke at most finitely many of those axioms. On the other hand SMAH+ is ZFC plus "for each \(k\), there exists a \(k\)-Mahlo cardinal". A proof could invoke the axiom to know that there exist \(k\)-Mahlo cardinals for arbitrarily high \(k\). :Another interesting example would be EFAE0 = EFA + for each \(k\), an axiom of the form "\(\omega \uparrow\uparrow k\) is well-ordered", versus EFAE0+ = EFA + "for each \(k\), \(\omega \uparrow\uparrow k\) is well-ordered". The former axioms are all provable in PA, so the former theory is equivalent to PA, but the last axiom is equivalent to "\(\varepsilon_0\) is well-ordered", which is not provable in PA, so EFAE0+ is stronger than PA. Goodstein's theorem is provahble in EFAE0+, but not EFAE0. Deedlit11 (talk) 04:55, July 29, 2014 (UTC) :I agree that this is quite confusing at first, I had the same problem when I first met the definitions on SMAH and SMAH+. The key thing about SMAH is that we don't allow logical operations on infinite number of statements, so we can't simply take union of infinitely many statements of form "there is \(k\)-Mahlo cardinal" to get "for all \(k\) there is \(k\)-Mahlo cardinal". :Deedlit's example also illustrates the fact on why there must be the least ordinal for which we can't prove it's well-ordered: at first one may think that since PA can prove that \(\omega\) is well ordered, \(\omega^\omega\) is, \(\omega^{\omega^\omega}\) is,... then, since all of these are provable in PA, it should be also the case that PA can prove that it can prove all of the above. One of the reasons why this isn't so obvious is because the proof of each fact requires much more time than earlier ones - so, possibly, while we can prove that \(\omega^\omega\) is well ordered in 500 symbols, for \(\omega^{\omega^\omega}\) we might require \(2\uparrow^{500} 500\). So only way would be to again use some infinitary operation, which is disallowed. LittlePeng9 (talk) 07:17, July 29, 2014 (UTC) First values of the F function I believe F(1)=0 or undefined (game must consist of n<1 stages, so that would be zero stages). I also think that F(2) = 1, but can we find more values or lower bounds? I don't understand it in full now. Wythagoras (talk) 08:09, July 29, 2014 (UTC) :Anyone? Wythagoras (talk) 16:33, August 21, 2014 (UTC) :If the game lasts zero stages, the statement "Bob has kept all his promises" is (vacuously) true, therefore Bob wins you're.so. 18:38, August 21, 2014 (UTC) Computability One thing confusing me here is the "for all T," "for all P and Q," "for all V and s." Doesn't that give us infinitely many things to evaluate? Friedman gives the three functions as functions of T, (P,Q), and (V,s) respectively. It appears that we have not three functions, but three infinite families. you're.so. 16:17, August 22, 2014 (UTC)